Question for proving something is bounded on an interval 
Let $f(x)$ = $x^2e^{-x}$. Show that $f$ is bounded on $(0,\infty)$

I know that $f$ is only bounded on the interval $I$ if and only if there exists some $M>0$ such that $|f(x)| <  M$ for all $x \in I$. But how would I go about translating this into a proof?
 A: You can use the Squeeze Theorem.  One possibility is:
$$\lim_{x \rightarrow \infty} e^{-x} \leq \lim_{x \rightarrow \infty} x^2e^{-x} \leq \lim_{x \rightarrow \infty} e^{x/2}e^{-x}.$$
Note that you can easily show that the upper and lower bound limits are both equal to zero.
You can argue that because your function is continuous, it is bounded on any closed interval $[0,N]$,$N \in \mathbb{R}$.
A: Without using derivatives.
We know that the function is bounded below by $0$. We know that $\ln(x)<\dfrac{x}{2}$ on the interval $(0,1)$ since $\ln(x)<0$ on that interval.
In the following animated GIF the curve is the graph of $y=\dfrac{1}{x}$, the dark shaded region has area $\ln(x)$ for $x\ge1$ and the orange rectangle upon which it is superimposed has area $\dfrac{x}{2}$.
This constitutes a "proof without words" that
$$ \ln(x)<\frac{x}{2}\quad \text{for }x\ge1$$

Thus we have that $$x^2<e^x\quad\text{for }x>0$$
Multiplying both sides of the inequality by the positive expression $e^{-x}$ yields.
$$x^2e^{-x}<1\quad \text{for }x>0$$
A: One approach is to note $$\int_0^\infty x^2e^{-x}dx = \Gamma(3) < \infty$$ where $$\Gamma(z) = \int_0^\infty t^{z-1}e^{-t}dt$$ is the Gamma function. It's not true in general that integrable implies bounded (for instance, consider $\int_0^1 x^{-1/2}dx$), but in your case the function $x^2e^{-x}$ is continuous everywhere, and therefore bounded on all (compact) subsets $[0, R]$ for all $R > 0$, in particular in a neighbourhood of zero. The fact the integral converges gives you boundedness in a neighbourhood of infinity. Continuity gives you boundedness everywhere else. 
